1.3.2 Venn Diagrams

You represent all possibilities as being somewhere inside a rectangle. Consider one circle in the rectangle. As a simple example, think of the game of darts. You’re either pretty wasted or pretty bad, so you hit anywhere inside the box with equal probability.

But suppose you score if you hit anywhere inside the circle. Hitting inside the circle means event $A$ has happened. Suppose the box has area $1$ (all area units here are $f{t}^2$), and the circle has an area of $.2$. Then the probability you’ll score (i.e. be inside the circle is) $.2/1=.2$. So you can say the probability of event $A$, that the dart lands inside the circle is $P(A)=.2$.

Now suppose you’re trying to make this a bit easier. You have another circle $B$ of area $.3$. You allow yourself to score when you get inside $B$ also.

If the circles are non-overlapping, then the probability of a score is $.2+.3=.5$. You can put this more succinctly by saying a score is when $A$ or $B$ occur. You can notate this as $A+B$. So this means that the probability of $A$ or $B$, $P(A+B)=P(A)+P(B)$, in this case (where $A$ and $B$ don’t overlap).

What happens if the circles $A$ and $B$ do overlap? The region of overlap you call the intersection of $A$ and $B$. This is often just denoted $AB$ like we did above. (You can roughly think of ”anding” as a multiplication.)

Logically that means the dart is both in A  and B . So the probability that you’ll throw the dart and it’ll land in this overlap region is the probability of A  and B . We denote this $P(AB)$. What’s that probability? In this example it depends on the the area of the overlap region $AB$. If it was $.1$ $P(AB)=.1$.

Now with this overlap, what’s the probability that you’ll score? It’d be wrong now to add .2 and .3, because you’re counting the areas of overlap twice. (Take the extreme case of A being totally inside B . Should get .3 in this case) So how do you handle this correctly? You want the total area that the two circles take up. When you add the areas of A and B   you’re counting the intersected area twice, not once like you should. So you better subtract that off to get the right area. So in this example $P(A+B)=.2+.3-.1=.4$.

You can then see that it makes sense to say in general: